pKa is 3,86 for lactic acid and at pH 2 there will be mostly free acid in you mixture.
If you have the mixture at pH 2, you can evaporate it (a lot of work, tedious to evaporate water) then you can dissolve the solid residue in MeOH, and filter this and evaporate again. Then you can wash this solid with EtOH and since calcium lactate is insoluble in EtOH, all impurities should be dissolved and the remaining solid should be calcium lactate (in the best scenario).
Then you can liberate your lactic acid with dilute sulfuric acid and the calcium sulphate formed will precipitete (its gypsum) and can be removed by filtration. Here you need to use exactly 2 equivalent sulphuric acid, otherwise you get lactic acid contaminetad with sulphuric acid Calcium lactate is really calcium dilactate. There is a risc that these filtations will be problematic because very fine particles are formed, so you can use Celite to facilitate the filtration. To extract a polar molecule like lactic acid from a bioorganic mixture like this is often problematic and takes long time, it would be best if you had a published procedure.